SEL-Ze drawn from the vertices of ABC to the opposite sides are equal,

1.	SEL-Ze drawn from the vertices of ABC to the opposite sides are equal, prove that the triangle is17, Ifthe altitudequilateral.
Answer» Given :AD = BE = CF and angles A = B = C= 90° To prove :Triangle ABC is equialteral. Proof : In triangle ABE and ACF Angle AEB = AFC ( 90° each ) BE = CF ( Given ) Angle A = A ( Common ) Hence Triangle ABE ~ ACF by AAS congruency AB = AC ( c.p.c.t ) In triangle AOE and EOC OE = OE ( Common ) Angle E = E ( 90° each ) AO = OC [ :. AD = FC, their halfs OA = OC ] Hence triangle AOE ~ EOC by RHS congruency AE = EC ( c.p.c.t ) In triangle ABE and BCE Angle E = E ( 90° each ) BE = BE ( common ) AE = EC ( proved above ) Hence, triangle ABE ~ BCE by SAS congruency. AB = BC ( c.p.c.t ) As AB = AC and AB = BC, so AC = BC. Hence, AB = BC = CA. HENCE PROVED.

SEL-Ze drawn from the vertices of ABC to the opposite sides are equal, prove that the triangle is17, Ifthe altitudequilateral.

Answer»

Given :AD = BE = CF and angles A = B = C= 90°

To prove :Triangle ABC is equialteral.

Proof : In triangle ABE and ACF

Angle AEB = AFC ( 90° each )

BE = CF ( Given )

Angle A = A ( Common )

Hence Triangle ABE ~ ACF by AAS congruency

AB = AC ( c.p.c.t )

In triangle AOE and EOC

OE = OE ( Common )

Angle E = E ( 90° each )

AO = OC [ :. AD = FC, their halfs OA = OC ]

Hence triangle AOE ~ EOC by RHS congruency

AE = EC ( c.p.c.t )

In triangle ABE and BCE

Angle E = E ( 90° each )

BE = BE ( common )

AE = EC ( proved above )

Hence, triangle ABE ~ BCE by SAS congruency.

AB = BC ( c.p.c.t )

As AB = AC and AB = BC, so AC = BC.

Hence, AB = BC = CA.

HENCE PROVED.